Can Any1 tell me why kienetic energy is equal to both m.v[SUP]2[/SUP] and 1/2

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Discussion Overview

The discussion revolves around the relationship between kinetic energy and its mathematical expressions, specifically why kinetic energy is represented as both \( m v^2 \) and \( \frac{1}{2} m v^2 \). Participants explore the derivation of these equations and the implications of work done in various scenarios, including a hypothetical situation involving a snooker player hitting a ball on a frictionless surface.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a derivation of kinetic energy as \( K.E = m v^2 \) using work-energy principles and Newton's equations of motion.
  • Another participant challenges the derivation by pointing out the incorrect assumption of constant velocity in the application of the equations.
  • A later reply clarifies that the average speed during acceleration should be considered, leading to the correct expression of kinetic energy as \( K.E = \frac{1}{2} m v^2 \).
  • In a separate inquiry, a participant questions the work done by a player hitting a snooker ball with a force of 6N, suggesting that the work done is zero despite energy being transferred to the ball.
  • Another participant argues against this notion, stating that a force cannot be exerted for zero time and that the model presented is flawed.
  • Further discussion introduces the concept of impulsive forces and how they can be applied over a very short time interval, affecting momentum and energy transfer.

Areas of Agreement / Disagreement

Participants express differing views on the derivation of kinetic energy and the implications of work done in the context of instantaneous forces. There is no consensus on the validity of the models presented, and multiple competing views remain throughout the discussion.

Contextual Notes

Participants highlight limitations in the assumptions made regarding constant velocity and the application of force over time. The discussion reveals dependencies on definitions and interpretations of force and work in dynamic scenarios.

Arslan
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Can Any1 tell me why kienetic energy is equal to both m.v2 and 1/2 m.v2


E=W=F.d
=m.a.d ......=mgh
=m.(v/t).d
=m.(d/t^2).d
=m.d^2/t^2
=m.v^2
So hence proved K.E= m.v^2



But 2.a.s=Vf2-Vi2 Newton equatio of motion
2.F/m.s=V2......if initial velocity is zero
F.s=1/2 mv2
W=E=1/2 mv2
So hence proved K.E=1/2 m.v^2
 
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Please don't double post

Arslan said:
=m.a.d
=m.(v/t).d
a=\frac{dv}{dt}\ne\frac{v}{t}
 


Arslan said:
E=W=F.d
=m.a.d ......=mgh
=m.(v/t).d

In the equation above, v is the final velocity at time t.

=m.(d/t^2).d

In this step you use v=d/t, but the object isn't moving at constant speed v. It starts with speed 0, and increases speed linearly until the end (v). d/t is the average speed which is v/2 for constant acceleration from v=0 to v=v. So v=2d/t.
 
Thanks i got it

i have still one more question related to energy

If a snooker player hits a ball on frictionless surface...he hits with force 6N and ball starts to move with constant velocity say 10m/s and never stops,
how much work is done by the player and how how much energy the ball has.

If we use W=integral(F.dx)
but F=6 at instant t=0 and F=0 for all other values of t
and d=0 at instant t=0 and d is increasing for all other values of t
So the product remains 0

It mean that work done is zero but energy is transferred to ball i.e. 1/2 mv^2
 
Arslan said:
If we use W=integral(F.dx)
but F=6 at instant t=0 and F=0 for all other values of t
and d=0 at instant t=0 and d is increasing for all other values of t
So the product remains 0

It mean that work done is zero but energy is transferred to ball i.e. 1/2 mv^2
No, it means that your model is flawed. You can't exert a force for zero time. ∫Fdt & ∫Fdx will both be greater than zero.
 
Doc Al said:
No, it means that your model is flawed. You can't exert a force for zero time.
Well you can conceptually exert a finite force for zero duration. It won't do anything, however.

You can also apply a rather large but finite force F(t) for a very small period of time δt. Taking ∫Fdt to be the change in momentum Δp and letting δt→0 so that Δp remains constant leads to the idea of impulsive forces. These can yield quite good estimates of the behavior if
  • You don't care about the deformations and such during that time interval δt during which the force is non-zero
  • The changes to the integrated state (position) during that time interval δt are very small
  • That large but finite force F(t) overwhelms all other forces during that time interval.
 
D H said:
Well you can conceptually exert a finite force for zero duration. It won't do anything, however.
True. :wink: